Download 2. Base For the Zariski Topology of Spectrum of a ring Let X be a

2. Base For the Zariski Topology of Spectrum of a ring
Let X be a topological space. A family of open sets B is a base for the topology on X
if every open set of X is a union of elements of B.
Proposition 2.1. The family of basic open sets B = {D(f ) : f ∈ A} forms a base for the
Zariski topology on Spec A.
Proof. Let U be an open subset of Spec A. Then there exists an ideal I of A so that
U = Spec A \ V (I). For each x ∈ U, we know x 6∈ V (I). In other words, I is not contained
in x. Then there exists fx ∈ I \ x. Hence fx 6∈ x and thus x 6∈ V (fx ). It is equivalent to say
that x ∈ D(fx ).
On the S
other hand, (fx ) ⊂ I and hence V (I) ⊂SV (fx ). We see that D(fx ) ⊂ U. Notice
that U ⊂ x∈U D(fx ) ⊂ U. We conclude that U = x∈U D(fx ). This shows that every open
set of X is a union of elements of B. By definition, B forms a basis for the Zariski topology
on Spec A.
Definition 2.1. A topological space is quasi compact if and only if every open covering
has a finite sub cover.
Proposition 2.2. The spectrum Spec A of a ring is quasi-compact.
Proof. Let {Uλ } be an open covering of X = Spec A. Since {D(f ) : f ∈ A} forms a base
for the Zariski topology
of X, we can in fact assume
PD(fλ ) for some
T that Uλ is of the form
S
V
(f
)
=
∅,
i.e.
V
(
D(f
)
=
X.
This
implies
that
fλ ∈ A.
Then
λ
λ
λ (fλ )) = V (1).
λ
λ
pP
pP
P
n ∈
(f
)
=
(1).
Hence
1
∈
(f
).
By
definition,
1
(f
)
for
some n. But
Hence
λ λ
λ λ
λ λ
P
n
aλn ∈ A and fλ1 , · · · , fλn so
1 =P
1. We see that λ (fλ ) = (1). We can choose aλ1 , · · · , P
that S ni=1 aλi fλi = 1. This implies that the unit ideal (1) = ni=1 (fλi ). This implies that
X = ni=1 D(fλi ). Hence Spec A is quasi-compact.
Corollary 2.1. Suppose A is a ring and f ∈ A. Then D(f ) is also quasi-compact.
Proof. We will prove that D(f ) is homeomorphic to Spec Af , where Af is the localization
Sf−1 A, where Sf = {f n : n ≥ 0}. Hence D(f ) is the spectrum of a ring. By the previous
proposition, D(f ) is quasi-compact.
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